Ask your own question, for FREE!
Mathematics 17 Online
OpenStudy (anonymous):

Who wants a medal? Can you teach me about series and sequences?

OpenStudy (anonymous):

Taking the Calc BC test tomorrow? Anyway, I can probably say a few things. Firstly, Taylor Series: You can generate a taylor series centered at x = a for a function by f(x) + f'(x)(x-a) + f''(x)(x-a)^2/2! +... I think you get the idea. Usually these use f(0), f'(0), etc. but sometimes they will specify a different place to do it.

OpenStudy (anonymous):

Now, on the note of convergence: The ratio test is your friend. Most convergence things can be done with by dividing the n+1th term by the nth term of the series. The series converges where the absolute value of that ratio < 1. Conditional convergence can occur where it = 1, you have to check for that other ways.

OpenStudy (anonymous):

Are you talking about Arithmetic Progression, Geometric Progression, Harmonic Progression or the other series that Jamal is giving here? @gandalfwiz

OpenStudy (anonymous):

1/(1-x) = 1 + x + x^2 + x^3 + ... + x^n where x < 1. You can substitute in (-x) for X to get 1/(1+x) = 1 - x + x^2 - ... + (-x)^n You can also integrate that series to get the ln(1+x) or -ln(1-x) depending on which one you integrate.

OpenStudy (anonymous):

The series for sin(x) = X- X^3/3! + X^5/5! - ... + (-1)^nx^2n+1/(2n+1)! You can get the cosine series by taking the derivative of that or by using the taylor series formula I poster earlier. That's how you can derive the sin series if you can forget it.

OpenStudy (anonymous):

The e^x series is just 1 + x + x^2/2! + ... +x^n/n!. Easy. Again, you can use the taylor series formula to get this.

OpenStudy (anonymous):

Oh, I forgot to mention that the ratio test is the limit as n goes to infinity of the ratio I mentioned earlier. Stupid me.

OpenStudy (anonymous):

Wait, before you go on Jamal, I'm not quite that advanced.

OpenStudy (anonymous):

Sorry to let you go on for so long, but the site just let me load my question-- this is just for precalc.

OpenStudy (anonymous):

Oh, lol, sorry. I just have the calc BC test tomorrow and I was studying this. Anything in particular I can help with? I'm not great at precalc, though.

OpenStudy (anonymous):

Haha, I've got a lot of friends taking that tomorrow too. I was watching Khan academy and he filled in the questions I had. But thanks anyways! You still get a medal :)

Can't find your answer? Make a FREE account and ask your own questions, OR help others and earn volunteer hours!

Join our real-time social learning platform and learn together with your friends!
Can't find your answer? Make a FREE account and ask your own questions, OR help others and earn volunteer hours!

Join our real-time social learning platform and learn together with your friends!